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anonymous

  • one year ago

A car of mass 1.5 x 103 kg is hoisted on the large cylinder of a hydraulic press. The area of the large piston is 0.20 m2, and the area of the small piston is 0.015 m2. a.Calculate the magnitude of the force of the small piston needed to raise the car with slow speed on the large piston. b.Calculate the pressure, in pascals and kilopascals, in this hydraulic press.

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. anonymous
    • one year ago
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    Hey there how are you?

  3. anonymous
    • one year ago
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    I hope all is well with you;)

  4. Michele_Laino
    • one year ago
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    yes! I'm fine! thanks! :)

  5. Michele_Laino
    • one year ago
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    here you have to keep in mind the principle of Pascal, namely the pressure of the fluid is the same on both pistons

  6. anonymous
    • one year ago
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    Where water is used as multiplier of force in call cases.

  7. anonymous
    • one year ago
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    Since for question a I know the area of small piston

  8. anonymous
    • one year ago
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    I guess all I need is divide the mass to be lifted by the area of the piston?

  9. Michele_Laino
    • one year ago
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    more precisely are used special oils, instead of water. the pressure exerted by the fluid on larger piston is: \[\Large p = \frac{{Mg}}{S} = \frac{{1500 \cdot 9.81}}{{0.2}} = ...Pascals\]

  10. anonymous
    • one year ago
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    Right... I needed to calculate the normal force acting on the object divided by the area of the piston.

  11. Michele_Laino
    • one year ago
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    that's right!

  12. anonymous
    • one year ago
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    According to my calculator the pascals required on the piston is 73500 pascals

  13. anonymous
    • one year ago
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    So that's considering if we only use the large piston right?

  14. Michele_Laino
    • one year ago
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    I got p = 73.6 kPa approximately

  15. anonymous
    • one year ago
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    On the other hand if I use the smaller piston the answer would be slightly greater I think

  16. anonymous
    • one year ago
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    I got p=981000 on my side for the smaller piston.

  17. Michele_Laino
    • one year ago
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    no, the pressure is the same through all the fluid. The force which is acting on the smaller piston, is perpendicular with respect to that piston, and its magnitude is: \[\Large F = p \times s = 73575 \cdot 0.015 = ...Newtons\]

  18. anonymous
    • one year ago
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    Oh ok.. but why are we only using the pressure applied to the large piston? Is it because of the structure of hydraulic press?

  19. Michele_Laino
    • one year ago
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    no, we are using the pression of the fluid which is transmitted by the fluid itself, namely your hydraulic press can be represented as below: |dw:1439828282269:dw|

  20. Michele_Laino
    • one year ago
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    what changes are the magnitude of the forces applied to both pistons

  21. Michele_Laino
    • one year ago
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    magnitudes*

  22. anonymous
    • one year ago
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    So basically a force is applied to one end and that results in other end

  23. Michele_Laino
    • one year ago
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    yes! we apply a little force and the hydraulic press multiply this little force

  24. Michele_Laino
    • one year ago
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    multiplies*

  25. anonymous
    • one year ago
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    73575 Pa on the large piston.

  26. anonymous
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    I understand until that point

  27. anonymous
    • one year ago
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    But then, isn't it like by 0.2/0.015 that the pressure is multiplied?

  28. Michele_Laino
    • one year ago
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    that pressure is acting inside all the fluid, so the pressure of 73575 is acting on both pistons

  29. Michele_Laino
    • one year ago
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    the pressure is the same!

  30. anonymous
    • one year ago
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    Are you saying that fluid supplied from the inside the hydraulic press?

  31. Michele_Laino
    • one year ago
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    in your exercise, we apply a force whose magnitude is 1,104 Newtons, and we can rise a car whose weight is: 14,715 Newtons

  32. Michele_Laino
    • one year ago
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    yes! the fluid is working inside the hydraulic press

  33. anonymous
    • one year ago
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    The force the small piston needs to lift the car right? so I need to apply force on the small piston to lift the car to the larger piston and force multiplies

  34. Michele_Laino
    • one year ago
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    correct! we can see a hydraulic press as a multiplier of force

  35. anonymous
    • one year ago
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    So it's correct to say 73575 Pa* 0.015= pressure require to lift the car? The only thing I don't understand is why we multiply by 0.015 for a pressure calculated using 0.2. Isn't it like we have to multiply by 0.015/0.2 ?

  36. Michele_Laino
    • one year ago
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    pressure is 73575 Pa, and 73575*0.015 is a force

  37. anonymous
    • one year ago
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    likewise 73575 Pa*0.2 is the force I need to apply to larger piston to do the same trick?

  38. Michele_Laino
    • one year ago
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    better is: 73575*0.2 is the force which is acting on the larger piston

  39. anonymous
    • one year ago
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    wait a moment it's taking long so

  40. anonymous
    • one year ago
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    that's my thanks for being patient with me XD

  41. Michele_Laino
    • one year ago
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    thanks! :)

  42. anonymous
    • one year ago
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    I am trying my best to understand

  43. Michele_Laino
    • one year ago
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    ok! :)

  44. anonymous
    • one year ago
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    You previously calculated the pressure applied to the large piston using the formula Mg/s=pascals. What if you used 0.015 instead of the 0.2? The overall pressure in the hydraulic press would then increase no?

  45. anonymous
    • one year ago
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    According to your reasoning that pressure is the same across all points in a hydraulic press, then the pressure calculated using 0.015 instead of 0.2 would be greater...

  46. Michele_Laino
    • one year ago
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    I have used 0.2 since 0.2 is the area of the piston on which the weight of the car is acting on

  47. Michele_Laino
    • one year ago
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    I can not use 0.015, since it will be an error

  48. Michele_Laino
    • one year ago
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    |dw:1439829785001:dw|

  49. Michele_Laino
    • one year ago
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    what I can know is the pressure transmitted by the fluid through all the hydraulic press, and that pressure, by definition, is: \[\Large p = \frac{{Mg}}{S} = \frac{{1500 \cdot 9.81}}{{0.2}}\]

  50. anonymous
    • one year ago
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    Oh I see then in terms of the larger piston the pressure acting on it is calculated by Mg/0.2. Right the pressure acting on the larger piston is accounted for by that. On the other hand, when I need to apply the pressure exerted by the car on larger piston, multiplying the previous pascals with 0.15 will be in accord with the principles of multiplication achieved by water.

  51. anonymous
    • one year ago
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    So multiplying that number with 0.015 will give the force required to lift the car although at a slower rate.

  52. anonymous
    • one year ago
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    Am I on the right line of reasoning?

  53. Michele_Laino
    • one year ago
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    yes! if I multiply the pressure by the area of the smaller piston, I get the force (magnitude) which I have to apply to that smaller piston, in order to raise the car

  54. Michele_Laino
    • one year ago
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    the concept is that pressure is the same in all points of the hydraulic press Whereas what is different are the magnitude of the forces which are acting on both pistons.

  55. Michele_Laino
    • one year ago
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    magnitudes*

  56. anonymous
    • one year ago
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    Right.. so in that case water is used as multiplier of force to lift a really heavy piece of object.

  57. Michele_Laino
    • one year ago
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    that's right!

  58. anonymous
    • one year ago
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    Yet the pressure acting in the water is same across all objects. Is it correct to say that water is dispersing the required force all through the water hence reducing the force required to lift the vehicle?

  59. anonymous
    • one year ago
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    this is such a fascinating concept.. Quite the counterintuitive nature of our nature.

  60. Michele_Laino
    • one year ago
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    I think that it is correct to say that the effect of a force, is the production of the correspondig pressure inside the fluid which is contacting that force, and in turn the effect of a pressure of a fluid, is the force produced on all the surfaces which are contacting that fluid

  61. Michele_Laino
    • one year ago
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    and cause-effect, namely force-pressure, are descrined by the subsequent formula: \[\Large P = \frac{F}{S}\]

  62. anonymous
    • one year ago
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    And depending on the area of surface that force induced by the object contacting the surface leading to space filled with liquids without leaks connected to another surface counteracting the force will be all dependent on the area upon which the respective forces are taking effect.

  63. anonymous
    • one year ago
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    So within the given pressure, greater surface implies more force required to meet the same pressure, and less surface means less force required to meet that pressure.

  64. Michele_Laino
    • one year ago
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    no, greater is the surface and greater is the forces, and less is the surface and less is the force, since the ratio force/surface has to be constant. The magnitude of the resultant forces, or of the applied forces depends on the area of the surfaces contacting those forces

  65. anonymous
    • one year ago
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    Yeah I realized I said the wrong

  66. anonymous
    • one year ago
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    So I deleted it

  67. Michele_Laino
    • one year ago
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    ok! your last reply is correct!

  68. anonymous
    • one year ago
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    Isn't this counterintuitive?

  69. anonymous
    • one year ago
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    Well, I guess the expanse of liquid present underneath the surface with greater area will make sense that corresponding force required induce such pressure is indeed greater

  70. anonymous
    • one year ago
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    On the other hand smaller surface with smaller liquid underneath would make it less force required the meet the pressure although distributed throughout all the fluid bodies.

  71. Michele_Laino
    • one year ago
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    I don't think, I think that it is intuitive, the problem is that we have no experience, in our daily life, about expansion of fluid

  72. Michele_Laino
    • one year ago
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    yes! exact!

  73. anonymous
    • one year ago
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    Yes! Yes! lack of experience with interaction of liquid makes it harder for us to visualize.

  74. Michele_Laino
    • one year ago
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    yes! and the best way is to use our imagination. As Einstein said: "Imagination is more important than knowledge"

  75. anonymous
    • one year ago
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    On the other hand, studying physics makes you capable of physically achieving anything if you know how to properly coordinate your muscles to execute the proper force required to achieve that phenomena.

  76. anonymous
    • one year ago
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    I think, in my personal opinion, all children should be taught the principles of physics so as to excel as well as increase "efficiency" in life. So much are wasted by ignorance

  77. anonymous
    • one year ago
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    I think too, that physics offers us a exemplary thinking mechanism.

  78. anonymous
    • one year ago
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    Also physics extends our lifespan

  79. anonymous
    • one year ago
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    By creating new neurons in your brain

  80. Michele_Laino
    • one year ago
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    I think that there is a time for study physics, what is important for children is play, since by means of play, all children can collect those experience which will allow to think in terms of our imagination

  81. anonymous
    • one year ago
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    True, from those natural experience will help us to relate physics to our past times. Richer the past times, greater the utility for physics

  82. Michele_Laino
    • one year ago
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    correct!

  83. anonymous
    • one year ago
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    And physics makes us nostalgic of our past memories.

  84. Michele_Laino
    • one year ago
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    yes!

  85. anonymous
    • one year ago
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    And ultimately makes life worthwhile. Physics makes all my struggles disappear.

  86. Michele_Laino
    • one year ago
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    throw a rock in the air, or running through a meadow, are all important things

  87. anonymous
    • one year ago
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    Physics is the best prescription for eliminating societal prejudice and personal stigmas for it provides us more objective insight upon which to rely all of the issues we encounter in life. This makes it easy for us to say that someone is wrong in saying something or the applications of medical physics such as how our neutrons interact would invent better ways to learn

  88. anonymous
    • one year ago
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    it's all physics after all our world is made of. And contemporary education has so far failed to quantify the most effective means by which to learn, hence resulting in "learning something until it finally sinks in" or just relying on our intuition rather than physics of neurology.

  89. anonymous
    • one year ago
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    that's why I love physics .

  90. Michele_Laino
    • one year ago
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    :)

  91. anonymous
    • one year ago
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    I bet you were on the same road when you were younger

  92. Michele_Laino
    • one year ago
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    not exactly, when I was younger, I had the concern, to give a logical accommodation to everything around me

  93. anonymous
    • one year ago
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    Oh I see, trying to explain mathematically everything around you o

  94. Michele_Laino
    • one year ago
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    yes! a logical or rational explanation

  95. Michele_Laino
    • one year ago
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    ok! when you need help in physics, keep in mind that I always will help you! :)

  96. anonymous
    • one year ago
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    Thanks so much Michele!

  97. Michele_Laino
    • one year ago
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    :)

  98. anonymous
    • one year ago
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    Do you mind proof reading another application from this question that I learned?

  99. anonymous
    • one year ago
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    2. I calculated the pressure acting on the floor when standing on both feet. The areas of my shoes were approximated to be 15cm^2 each, making ti 30cm^2. Given that I weigh around 58kg, and pressure is calculated as P=Mg/s (surface area) P=58kg*9.8m/s^2/0.003m^3=189466.667 The pressure acting on the floor when standing on both feet is 1.9*10^5 Pa

  100. anonymous
    • one year ago
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    According to your P=Mg/s I think my reasoning above seems pretty good

  101. Michele_Laino
    • one year ago
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    yes! Your reasoning is correct!

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